Proving a Trigonometric Identity Using Exponential Functions

[DrKareem]

Hi. I need to prove the following identity

\arccos{z} =i \ln { z + (z^2 -1)^\frac{1}{2} }

I was given a hint to write
\cos{A}=z,
then rewrite
\cos{A}

in terms of the exponential.
\cos{A}=\frac{\exp{iA}+\exp{-iA}}{2}=z

I took the log on both sides and got stuck at that point.

\ln{\exp{iA} + \exp{-iA}}=\ln{z^2}

I know it's a correct method becaues the right hand is starting to take form. But i just couldn't solve for A (which will be \arccos{z} right?).

 

[Hurkyl]

Why take a log at that point? It's essentially a quadratic equation.

 

[Gib Z]

I have a favorite way of proving equation :D

say we have u=v

I let f(x)=u/v, Prove that the derivative is zero, so the answer is unchanging. Then I solve it for 1 value, we know u/v=k, k is some value. Fun Fun :)

In This case we could write f(z)=\frac{\arccos{z}}{i \ln { z + (z^2 -1)^\frac{1}{2} }}. Find f'(z), as you will see it is equal to zero. Then prove it for one value of z, 1 looks like the easies to me :). You will see f(z)=1. Multiply both sides by the denominator and its all good :)

 

[DrKareem]

Why take a log at that point? It's essentially a quadratic equation.

There must be something trivial that I'm not seeing. I still can't figure out how to solve for A.

I have a favorite way of proving equation :D

say we have u=v

I let f(x)=u/v, Prove that the derivative is zero, so the answer is unchanging. Then I solve it for 1 value, we know u/v=k, k is some value. Fun Fun :)

In This case we could write LaTeX graphic is being generated. Reload this page in a moment.. Find LaTeX graphic is being generated. Reload this page in a moment., as you will see it is equal to zero. Then prove it for one value of z, 1 looks like the easies to me :). You will see f(z)=1. Multiply both sides by the denominator and its all good :)

1) I didn't quite understand your method
2) I prefer deriving rather than proving it.

Thanks so far for the help.

 

[Hurkyl]

What if A were -i ln B?

 

[Mute]

in terms of the exponential.
\cos{A}=\frac{\exp{iA}+\exp{-iA}}{2}=z

I took the log on both sides and got stuck at that point.

\ln{\exp{iA} + \exp{-iA}}=\ln{z^2}

I know it's a correct method becaues the right hand is starting to take form. But i just couldn't solve for A (which will be \arccos{z} right?).

For one thing, you didn't take the log correctly to begin with - you would have gotten ln(2z), not z^2, but that's irrelevant, because as Hurkyl said you don't need to take a logarithm. Instead, just get everything to one side:

e^{Ai} + e^{-Ai} - 2z = 0

Now then, what if you multiplied the equation by e^{Ai}? Connect the result of this multiplication with Hurkyl's first comment; can you see then how to proceed?

 

[Gib Z]

Well basically, to understand my method, say we wish to prove u=v.

If u=v, then y=u/v is constant, 1. y is always equal to 1 is u and v are equal. So, to show the y value is never changing, we take the derivative of u/v. The derivative is basically the rate of change. If the derivative is zero, then the value never changes. So, we just take the derivative, prove it is zero. Now we know the value never changes, solve u/v for 1 easy value, in this case 1.

Now we know u/v=1, or u=v.

Of course, as you said, this is a proof rather than a derivation. But you asked for a proof, I gave you one :P

 

[DrKareem]

Thanks a lot for your help, I was able to solve it.